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Manager  G
Joined: 03 Jun 2019
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The outer dimensions of a closed rectangular cardboard box are 8 centi  [#permalink]

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35 00:00

Difficulty:   95% (hard)

Question Stats: 36% (02:43) correct 64% (02:34) wrong based on 359 sessions

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The outer dimensions of a closed rectangular cardboard box are 8 centimeters by 10 centimeters by 12 centimeters, and the six sides of the box are uniformly 1/2 centimeter thick. A closed canister in the shape of a right circular cylinder is to be placed inside the box so that it stands upright when the box rests on one of its sides. Of all such canisters that would fit, what is the outer radius, in centimeters, of the canister that occupies the maximum volume?

A. 3.5
B. 4
C. 4.5
D. 5
E. 5.5

PS56710.02
Manager  P
Joined: 14 Apr 2017
Posts: 79
Location: Hungary
GMAT 1: 760 Q50 V42 WE: Education (Education)
The outer dimensions of a closed rectangular cardboard box are 8 centi  [#permalink]

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parkhydel wrote:
The outer dimensions of a closed rectangular cardboard box are 8 centimeters by 10 centimeters by 12 centimeters, and the six sides of the box are uniformly 1/2 centimeter thick. A closed canister in the shape of a right circular cylinder is to be placed inside the box so that it stands upright when the box rests on one of its sides. Of all such canisters that would fit, what is the outer radius, in centimeters, of the canister that occupies the maximum volume?

A. 3.5
B. 4
C. 4.5
D. 5
E. 5.5

PS56710.02

The inner dimensions of the rectangular box are 7 x 9 x 11 because we have to subtract the side thickness twice from each outer dimension.

Case 1: The dimensions of the base of the box are 7 x 9.

In this case, the maximum radius for the base of the cylinder is 3.5, and consequently the volume of the cylinder is

V = (3.5)^2(pi)(11) = 134.75(pi)

Case 2: The dimensions of the base of the box are 7 x 11.

In this case, the maximum radius for the base of the cylinder is again 3.5.

V = (3.5)^2(pi)(9) < 134.75(pi)

Case 3: The dimensions of the base of the box are 9 x 11.

In this case, the maximum radius for the base of the cylinder is 4.5.

V = (4.5)^2(pi)(7) = (20.25)(pi)(7) > 134.75(pi)

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Originally posted by ZoltanBP on 28 Apr 2020, 10:52.
Last edited by ZoltanBP on 03 May 2020, 03:11, edited 1 time in total.
##### General Discussion
Intern  B
Joined: 11 Nov 2018
Posts: 13
Re: The outer dimensions of a closed rectangular cardboard box are 8 centi  [#permalink]

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ZoltanBP wrote:
parkhydel wrote:
The outer dimensions of a closed rectangular cardboard box are 8 centimeters by 10 centimeters by 12 centimeters, and the six sides of the box are uniformly 1/2 centimeter thick. A closed canister in the shape of a right circular cylinder is to be placed inside the box so that it stands upright when the box rests on one of its sides. Of all such canisters that would fit, what is the outer radius, in centimeters, of the canister that occupies the maximum volume?

A. 3.5
B. 4
C. 4.5
D. 5
E. 5.5

PS56710.02

The inner dimensions of the rectangular box are 7 x 9 x 11 because we have to subtract the side thickness twice from each outer dimension.

Case 1: The dimensions of the base of the box are 7 x 9.

In this case, the maximum radius for the base of the cylinder is 3.5, and consequently the volume of the cylinder is

V = (3.5)^2(pi)(11) = 134.75

Case 2: The dimensions of the base of the box are 7 x 11.

In this case, the maximum radius for the base of the cylinder is again 3.5.

V = (3.5)^2(pi)(9) < 134.75

Case 3: The dimensions of the base of the box are 9 x 11.

In this case, the maximum radius for the base of the cylinder is 4.5.

V = (4.5)^2(pi)(7) = (20.25)(pi)(7) > 134.75

Why can't 11 be the base. Then the radius will be 5.5
Manager  P
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Re: The outer dimensions of a closed rectangular cardboard box are 8 centi  [#permalink]

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PrajwalGupta wrote:
Why can't 11 be the base. Then the radius will be 5.5

If one dimension of the inner rectangular base of the box is 11, then the other dimension must be either 7 or 9. If the radius of the base of the cylinder is 5.5, then the cylinder cannot be placed inside the box to stand on that base because the rectangular box would be too narrow to contain the cylinder.
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Intern  B
Joined: 11 Nov 2018
Posts: 13
Re: The outer dimensions of a closed rectangular cardboard box are 8 centi  [#permalink]

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ZoltanBP wrote:
PrajwalGupta wrote:
Why can't 11 be the base. Then the radius will be 5.5

If one dimension of the inner rectangular base of the box is 11, then the other dimension must be either 7 or 9. If the radius of the base of the cylinder is 5.5, then the cylinder cannot be placed inside the box to stand on that base because the rectangular box would be too narrow to contain the cylinder.

Got it, sir. I overlooked that.

Thank you for the explanation
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The outer dimensions of a closed rectangular cardboard box are 8 centi  [#permalink]

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parkhydel wrote:
The outer dimensions of a closed rectangular cardboard box are 8 centimeters by 10 centimeters by 12 centimeters, and the six sides of the box are uniformly 1/2 centimeter thick. A closed canister in the shape of a right circular cylinder is to be placed inside the box so that it stands upright when the box rests on one of its sides. Of all such canisters that would fit, what is the outer radius, in centimeters, of the canister that occupies the maximum volume?

A. 3.5
B. 4
C. 4.5
D. 5
E. 5.5

PS56710.02

Dimensions available inside of the rectangular cardboard box are (8 - $$\frac{1}{2}$$ - $$\frac{1}{2}$$), (10 - $$\frac{1}{2}$$ - $$\frac{1}{2}$$) and (12 - $$\frac{1}{2}$$ - $$\frac{1}{2}$$) i.e. 7, 9 and 11

Volume of right circular cylinder = $$π * r^2 * h$$. Here, volume is a function of 'r' and 'h'.

Here if the effect of an increase in r is more than increase in h if increase is same for both. So, for Volume to be maximum the guiding parameter is radius 'r' beause degree of 'r' is more than degree of 'h'.

Possible dimension of 'r' an be chosen from three rectangular surfaces with dimensions 7,9; 9,11 and 7,11, lower dimension deciding the final dimension.
So, possible values of 'r' are $$\frac{7}{2}$$, $$\frac{9}{2}$$ and $$\frac{7}{2}$$ and heights would be 11, 7 and 9 respectively.

But verifying it since height would differ in the three cases.
Volume for if h is 11 = $$π * (\frac{7}{2})^2 * 11 = \frac{439}{4}π$$

Volume for if h is 7 = $$π * (\frac{9}{2})^2 * 7 = \frac{567}{4}π$$

Volume for if h is 9 = $$π * (\frac{7}{2})^2 * 9 = \frac{441}{4}π$$

Hence required radius = $$\frac{9}{2}$$ = 4.5

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Originally posted by unraveled on 07 May 2020, 03:36.
Last edited by unraveled on 07 May 2020, 04:39, edited 1 time in total.
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Location: India
Re: The outer dimensions of a closed rectangular cardboard box are 8 centi  [#permalink]

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Hey lnm87

Your answer would be wrong, if the outer dimensions of a closed rectangular cardboard box were 8 centimeters by 10 centimeters by 13 centimeters. (just a slight change)

Volume depends on $$d^2*h$$

2 cases possible-

1. $$d=9; (d^2*h)_{max}= 9^2*7 = 81*7$$

2.$$d=7; (d^2*h)_{max}= 7^2*12 = 84*7$$

Sloan MIT School Moderator V
Joined: 07 Mar 2019
Posts: 1299
Location: India
GMAT 1: 580 Q43 V27
WE: Sales (Energy and Utilities)
The outer dimensions of a closed rectangular cardboard box are 8 centi  [#permalink]

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nick1816 wrote:
Hey lnm87

Your answer would be wrong, if the outer dimensions of a closed rectangular cardboard box were 8 centimeters by 10 centimeters by 13 centimeters. (just a slight change)

Volume depends on $$d^2*h$$

2 cases possible-

1. $$d=9; (d^2*h)_{max}= 9^2*7 = 81*7$$

2.$$d=7; (d^2*h)_{max}= 7^2*12 = 84*7$$

Ahh...!!
Missed the height part. Thanks nick for pointing out, I will rectify it.
I think you meant 11.
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Posts: 9
Re: The outer dimensions of a closed rectangular cardboard box are 8 centi  [#permalink]

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Let me make it very simple, it asked for radius of max. cylinder the box can contain.
Now the max volume of cylinder is close to the total volume of box.

So equate 10*8*12>pi*r^2*h
h=12

10*8>π*r^2
r^2<24
Approx . 4.5

Posted from my mobile device
Manager  B
Joined: 12 Nov 2018
Posts: 98
Re: The outer dimensions of a closed rectangular cardboard box are 8 centi  [#permalink]

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is this in the official OG?

isn't it too tough for GMAT?
Math Expert V
Joined: 02 Sep 2009
Posts: 65290
Re: The outer dimensions of a closed rectangular cardboard box are 8 centi  [#permalink]

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prototypevenom wrote:
is this in the official OG?

isn't it too tough for GMAT?

Yes. Please check the tags: Source: OG 2021 and Source: Official Guide.
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Re: The outer dimensions of a closed rectangular cardboard box are 8 centi  [#permalink]

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parkhydel wrote:
The outer dimensions of a closed rectangular cardboard box are 8 centimeters by 10 centimeters by 12 centimeters, and the six sides of the box are uniformly 1/2 centimeter thick. A closed canister in the shape of a right circular cylinder is to be placed inside the box so that it stands upright when the box rests on one of its sides. Of all such canisters that would fit, what is the outer radius, in centimeters, of the canister that occupies the maximum volume?

A. 3.5
B. 4
C. 4.5
D. 5
E. 5.5

PS56710.02

We see that each of the inner dimensions of the box is 2 x ½ = 1 cm less than its respective outer dimension. Therefore, the inner dimensions of the box are 7 cm by 9 cm by 11 cm. Since the canister has to be able to stand upright when the box rests on one of its sides, the maximum value if the canister’s radius is 4.5 cm (notice that would make the diameter = 9 cm, equaling the smaller of the dimensions of the base of the box if the base is 9 cm by 11 cm and height is 7 cm).

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